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2 years ago

ERROR ANALYSIS Describe and correct the error in

Mathematics

1 answer:

valkas [14]2 years ago

3 0

Answer:

The graph should be stretched rather than become narrower.

Step-by-step explanation:

To figure this out, just create some example points.

At x = 0, your y-value will always be 0. But if you were to plug in the value 1, you would get a y-value of 1 in y=x^2, but a value of 0.5 in y=0.5x^2. If you were to plug in a value of 2, you would get a value of 4 in y=x^2, but a value of 2 in y=0.5x^2.

If you continue this pattern for a few more points, then plot them, you will see that adding a coefficient of 0.5 simply stretches the graph

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What is -10a to the 6th power over -9 simplified

densk [106]

Answer: 5/3(a)^6

Step-by-step explanation:

8 0

3 years ago

Pls help me out ! i’ll mark you brainliest

Vlad [161]

If s=2, then t= 2(2)-3 = 1
T=2(4)-3= 5
T=2(6)-3= 9
T=2(8)-3=13

6 0

3 years ago

Read 2 more answers

The value of x varies directly with y, and whenX= 2/3, y = 6. Find the value of y when x = 10-24A.y=9B.y=4C. y=93D.y=1

uysha [10]

Problem

Solution

for this case we know that when x =2/3 , y= 6

$10\frac{1}{3}=\frac{10\cdot3+1}{3}=\frac{30+1}{3}=\frac{31}{3}$

Then we want to find the value of y= ? when x= 10 1/3 = 31/3 and we can do this:

6/(2/3) = y/(31/3)

$\frac{6}{\frac{2}{3}}=\frac{y}{\frac{31}{3}}$ $y=\frac{31}{3}\cdot\frac{6}{\frac{2}{3}}=\frac{31}{3}\cdot\frac{3}{2}\cdot\frac{6}{1}=\frac{31\cdot3\cdot6}{3\cdot2\cdot1}=93$

And solving for y we got:

y = (3/2) * 6 * (31/3) = 31*3 = 93

C. y= 93

7 0

1 year ago

The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean 1263 and a stan

DENIUS [597]

Answer:

(a) The 29th percentile for the number of chocolate chips in a bag is 1198.65.

(b) The number of chocolate chips in a bag that make up the middle 95% of bags are [1146, 1380].

Step-by-step explanation:

Let the random variable X represent the number of chocolate chips in a bag of chocolate chip cookies.

The random variable X is normally distributed with mean, μ = 1263 and a standard deviation, σ = 117.

(a)

Compute the 29th percentile for the number of chocolate chips in a bag as follows:

P (X < x) = 0.29

⇒ P (Z < z) = 0.29

The value of z for the above probability is, z = -0.55.

Compute the value of x as follows:

$z=\frac{x-\mu}{\sogma}\\-0.55=\frac{x-1263}{117}\\x=1263-(117\times 0.55)\\x=1198.65$

Thus, the 29th percentile for the number of chocolate chips in a bag is 1198.65.

(b)

According to the Empirical rule 95% of the normally distributed data lies within 2 standard deviations of the mean.

P (μ - σ < X < μ + σ) = 0.95

P (1263 - 117 < X < 1263 + 117) = 0.95

P (1146 < X < 1380) = 0.95

Thus, the number of chocolate chips in a bag that make up the middle 95% of bags are [1146, 1380].

(c)

The inter-quartile range of the normal distribution is:

IQR = 1.349 σ

Compute the inter-quartile range of the number of chocolate chips in a bag of chocolate chip cookies as follows:

IQR = 1.349 σ

= 1.349 × 117

= 157.833

Thus, the inter-quartile range of the number of chocolate chips in a bag of chocolate chip cookies is 157.83.

5 0

3 years ago

Hello chaps, please help me with this absurb eqution, "I think of a number add one then double the result

ziro4ka [17]

Answer:

2n+2

Step-by-step explanation:

Let the number = n.

Add 1 to n:

n+1

Double n+1:

2(n+1)

Distribute:

2n+2

6 0

3 years ago